ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A complete sum-check transcript over F_101

Example

Over $F_{101}$ let $g (X, Y, Z) = 2 X^{3} + X Z + Y Z$ , with individual bounds $(3, 1, 1)$ and claimed sum $H = 12$ . An honest transcript with successive challenges $2, 3, 6$ has messages $h_{1} (X) = 8 X^{3} + 2 X + 1, h_{2} (Y) = 34 + Y, h_{3} (Z) = 16 + 5 Z .$ The successive scalar claims are $69, 37, 46$ , and the terminal equality holds. For this polynomial, a false initial sum claim is accepted with probability at most $5 / 101$ by the randomized protocol; a displayed fixed challenge sequence alone is not a soundness proof.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

Honest messages are partial cube sums, with coefficient, endpoint-consistency and terminal checks (Sum-check with explicit degree bounds).

[F2]

For any false initial claim, acceptance is at most min(1,sum_i d_i/|F|) under fresh independent challenges and trusted evaluation (Adaptive-prover soundness of sum-check).

Verification

1.1

Sum $g$ over Boolean $Y, Z$ to obtain $q_{1} (X) = 8 X^{3} + 2 X + 1$ . Then $q_{1} (0) + q_{1} (1) = 1 + 11 = 12$ and $q_{1} (2) = 64 + 4 + 1 = 69$ . This also computes the cube sum by splitting over $X$ .

F1algebra

2.1

After $r_{1} = 2$ , sum $g (2, Y, Z)$ over $Z = 0, 1$ : $q_{2} (Y) = 16 + (16 + 2 + Y) = 34 + Y$ . Its endpoint sum is $34 + 35 = 69$ , and its value at $r_{2} = 3$ is $37$ .

F1step 1.1algebra

3.1

With that prefix $q_{3} (Z) = g (2, 3, Z) = 16 + 5 Z$ . Its endpoint sum is $16 + 21 = 37$ . At $r_{3} = 6$ its value is $46$ , also $g (2, 3, 6) = 16 + 12 + 18 = 46$ . The coefficient lists have lengths $4, 2, 2$ and the degrees meet $(3, 1, 1)$ , so all checks pass modulo 101.

F1step 2.1algebra

4.1

For any false initial claim the soundness theorem applies to the fixed polynomial and trusted degree bounds, giving error at most $(3 + 1 + 1) / 101 = 5 / 101$ over fresh independent challenges. It asserts this for all adaptive provers, separately from the honest transcript just computed.

F2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Justin Thaler, Proofs, Arguments, and Zero-Knowledge (2023), §4.1 Example Execution pp.36–37 (standard reference, not scraped)